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3 years ago

12

Which best describes the figure that is represented by the following coordinates?

1 answer:

kolezko [41]3 years ago

7 0

Answer: its square.

Step-by-step explanation:

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5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0

3 years ago

Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution

Blababa [14]

Answer:

If you mean only one rational solution, the answer is

$k_1 = 8, k_2 = -8$

If you mean at least 1 rational solution, the answer is

$k\in (-\infty, -8]\cup[8, \infty)$

Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

$\Delta = b^2 - 4ac$

$\Delta = k^2 -4 \cdot 4 \cdot 4$

$\Delta = k^2 -64$

Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

$k^2 - 64 > 0$

and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

$k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)$

$\boxed{\text{For } k^2 - 64 = 0}$

$k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8$

7 0

3 years ago

George Orwell was visiting London and left his hotel at 8:00 am and returned at 9:00 pm. That day, he was recorded 286 times by

cupoosta [38]

Orwell, on average, was photographed 22 times every hour.

4 0

3 years ago

Read 2 more answers

Determine the constant rate of change. please know what is before answering.

Nataly_w [17]

The constant rate of change is 82

7 0

3 years ago

5-x=3<br> Work out the value of 3x2

Butoxors [25]

Answer:

No

Step-by-step explanation:

5-X=3

X=5+3

X=8

5-8=3

6 0

3 years ago